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Unformatted text preview: Nguyen, Thanh – Homework 10 – Due: Oct 9 2007, 7:00 pm – Inst: D Weathers 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A 2 . 83 g bullet moving at 270 m / s penetrates a tree to a depth of 6 . 02 cm. Use energy considerations to find the aver age frictional force that stops the bullet. Correct answer: 1713 . 51 N. Explanation: We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 2 mv 2 = f · s Solving for the frictional force, f , f = mv 2 2 s = (0 . 00283 kg)(270 m / s) 2 2(0 . 0602 m) = 1713 . 51 N . 002 (part 2 of 2) 10 points Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped. Correct answer: 0 . 000445926 s. Explanation: If we assume the frictional force is constant, by Newton’s second law we can assume con stant deceleration, which implies v avg = v initial 2 . The time of deceleration is then t = s v avg = . 0602 m 270 m / s 2 = 0 . 000445926 s . keywords: 003 (part 1 of 1) 10 points A car of weight 2110 N operating at power 121 kW develops a maximum speed of 29 . 3 m / s on a level, horizontal road. Assume: The resistive force remains con stant under the conditions mentioned below. What is the maximum speed of the car up an incline of 5 . 51 ◦ relative to the horizontal? Correct answer: 27 . 9298 m / s. Explanation: Basic Concepts: P = dW dt = ~ F · ~v . Maximum Speed and Terminal Velocity Resistive forces. Let : W = 2110 N , p = 121 kW , v 1 = 29 . 3 m / s , and θ = 5 . 51 ◦ . Solution: On level ground, the only forces acting on the car in the horizontal direction are the applied force F 1 and the resistive force R . (Gravity and the normal force act perpen dicular to this and add vectorially to zero.) Since the car has reached it maximum speed, it is no longer accelerating, so the net force is zero. That is F 1 R = 0 so R = F 1 . The applied force is related to the power through the relation P = ~ F 1 · ~v 1 . The ap plied force, ~ F 1 , is in the same direction as the velocity, so ~ F 1 · ~v = F 1 v 1 . This means that F 1 = P v 1 = 121 kW 29 . 3 m / s µ 10 3 W kW ¶ = 4129 . 69 N . Nguyen, Thanh – Homework 10 – Due: Oct 9 2007, 7:00 pm – Inst: D Weathers 2 When the car is on the incline, we now have...
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 Winter '08
 Weathers
 Energy, Force, Potential Energy, Work, Correct Answer, Thanh

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